BOOK I.

PROBLEM I. (Fig. 1.)

In any indefinite ſtraight line, let the Point A be aſſigned; it is required

to cut it in ſome other point O, ſo that the ſquare on the ſegment AO

may be to the ſquare on a given line, P, in the ratio of two given ſtraight

lines R and S.

Analysis
. Since, by Hypotheſis, the ſquare on AO muſt be to the

ſquare on P as R is to S, the ſquare on AO will be to the Square on P as

the ſquare on R is to the rectangle contained by R and S (
Eu
. V. 15.) Let there be taken AD, a mean proportional between AB (R) and AC

(S); then the Square on AO is to the ſquare on P as the ſquare on R is

to the ſquare on AD, or (
Eu
. VI. 22) AO is to P as R to AD; conſe-

quently, AO is given by
Eu
. VI. 12.

Synthesis
. Make AB equal to R, AC equal to S, and deſcribe on

BC a ſemi-circle; erect at A the indefinite perpendicular AF, meeting the

circle in D, and take AF equal to P; draw DB, and parallel thereto FO,

meeting the indefinite line in O, the point required.

For, by reaſon of the ſimilar triangles ADB, AFO, AO is to AF (P) as

AB (R) is to AD; therefore (
Eu
. VI. 22.) the ſquare on AO is to the

ſquare on P as the ſquare on R is to the ſquare on AD; but the ſquare on

AD is equal to the rectangle contained by AB (R) and AC (S) by
Eu
. VI. 13. 17; and ſo the ſquare on AO is to the ſquare on P as the ſquare on R

is to the rectangle contained by R and S; that is (
Eu
. V. 15.) as R is to S.

Q. E. D.

Scholium
. Here are no limitations, nor any precautions whatever to be

obſerved, except that AB (R) muſt be ſet off from A that way which O

is required to fall.